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Difference between revisions of "2010 AMC 8 Problems/Problem 24"

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\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
  
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==Solution 1==
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== Solution 1==
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Use brute force.
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Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
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<math>10^8=100,000,000</math>,  
 
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<math>5^{12}=244,140,625</math>, and
 
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<math>2^{24}=16,777,216</math>.
 
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Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer. (Not recommended for the contest and will take forever)
 
  
 
== Solution 2==
 
== Solution 2==
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Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
 
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== Solution 3==
 
 
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
 
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
 
<math>10^8</math> is as fine as it is.
 
<math>10^8</math> is as fine as it is.
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Solution by Math
 
Solution by Math
  
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==Solution 4==
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==Solution 3==
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We know that <math>10^{8}=(5*2)^{8}=5^{8}*2^{8}</math>. We also know that <math>5^{12}=5^{(8+4)}=5^{8}*5^{4}</math>. If we remove the common factor of <math>5^{8}</math> from both expressions, we are left with <math>2^{8}</math>, which equals 256, and <math>5^{4}</math>, which equals 625. So we know that <math>5^{12}</math> is bigger than <math>10^{8}</math>. This leaves us with only 2 options, A or E. Now we need to figure out which is bigger, <math>10^{8}</math> or <math>2^{24}</math>. To do this, we rewrite <math>2^{24}</math> as <math>(2^{8})^{3}=(2^{3})^{8}=(8)^{8}</math>, which is clearly less than <math>10^{8}</math>. Therefore, <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
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We know that <math>10^{8}=({5}\cdot{2})^{8}=5^{8}\cdot2^{8}</math>. We also know that <math>5^{12}=5^{(8+4)}=5^{8}\cdot5^{4}</math>. If we remove the common factor of <math>5^{8}</math> from both expressions, we are left with <math>2^{8}</math>, which equals 256, and <math>5^{4}</math>, which equals 625. So we know that <math>5^{12}</math> is bigger than <math>10^{8}</math>. Now we need to figure out which is bigger, <math>10^{8}</math> or <math>2^{24}</math>. To do this, we rewrite <math>2^{24}</math> as <math>(2^{3})^{8}=(8)^{8}</math>, which is clearly less than <math>10^{8}</math>. Therefore, <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
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By naman14
 
By naman14
  

Latest revision as of 07:12, 9 November 2024

Problem

What is the correct ordering of the three numbers, $10^8$, $5^{12}$, and $2^{24}$?

$\textbf{(A)}\ 2^{24}<10^8<5^{12}\\ \textbf{(B)}\ 2^{24}<5^{12}<10^8 \\ \textbf{(C)}\ 5^{12}<2^{24}<10^8 \\ \textbf{(D)}\ 10^8<5^{12}<2^{24} \\ \textbf{(E)}\ 10^8<2^{24}<5^{12}$

Solution 1

Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get $10^2=100$, $5^3=125$, and $2^6=64$. Since $64<100<125$, it follows that $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer.

Solution 2

First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. $10^8$ is as fine as it is. We can rewrite $2^{24}$ as $(2^3)^8=8^8$. Then we can rewrite $5^{12}$ as $(5^{\frac{3}{2}})^8=(\sqrt{125})^8)$. We take the eighth root of all of these to get ${10, 8, \sqrt{125}}$. Obviously, $8<10<\sqrt{125}$, so the answer is $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$. Solution by Math

Solution 3

We know that $10^{8}=({5}\cdot{2})^{8}=5^{8}\cdot2^{8}$. We also know that $5^{12}=5^{(8+4)}=5^{8}\cdot5^{4}$. If we remove the common factor of $5^{8}$ from both expressions, we are left with $2^{8}$, which equals 256, and $5^{4}$, which equals 625. So we know that $5^{12}$ is bigger than $10^{8}$. Now we need to figure out which is bigger, $10^{8}$ or $2^{24}$. To do this, we rewrite $2^{24}$ as $(2^{3})^{8}=(8)^{8}$, which is clearly less than $10^{8}$. Therefore, $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer.

By naman14

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=381

Video by MathTalks

https://youtu.be/mSCQzmfdX-g


Video Solution by WhyMath

https://youtu.be/EfCyJF1FEO ~someone

See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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